Q99 / 150 · MHT CET 2026 (17 April Shift 1) · Mathematics
A box contains 8 red and N green balls. Two balls are drawn at random from it. If X is the random variable representing the number of green balls drawn and $E(X) = 1.2$, then $N = \ldots$
- A$4$
- B$8$
- Correct answerC$12$
- D$16$
Solution
Total number of balls $= N + 8$
Two balls are drawn at random. Let $X$ be the number of green balls drawn. The possible values of $X$ are $0, 1, 2$.
$P(X=1) = \dfrac{^{N}C_{1} \times ^{8}C_{1}}{^{N+8}C_{2}} = \dfrac{16N}{(N+8)(N+7)}$
$P(X=2) = \dfrac{^{N}C_{2}}{^{N+8}C_{2}} = \dfrac{N(N-1)}{(N+8)(N+7)}$
The expected value $E(X)$ is given by:
$E(X) = 1 \times P(X=1) + 2 \times P(X=2)$
$E(X) = \dfrac{16N + 2N(N-1)}{(N+8)(N+7)} = \dfrac{2N(N+7)}{(N+8)(N+7)} = \dfrac{2N}{N+8}$
Given $E(X) = 1.2$, we have:
$\dfrac{2N}{N+8} = 1.2$
$2N = 1.2N + 9.6$
$0.8N = 9.6$
$N = 12$
Answer: $12$
Two balls are drawn at random. Let $X$ be the number of green balls drawn. The possible values of $X$ are $0, 1, 2$.
$P(X=1) = \dfrac{^{N}C_{1} \times ^{8}C_{1}}{^{N+8}C_{2}} = \dfrac{16N}{(N+8)(N+7)}$
$P(X=2) = \dfrac{^{N}C_{2}}{^{N+8}C_{2}} = \dfrac{N(N-1)}{(N+8)(N+7)}$
The expected value $E(X)$ is given by:
$E(X) = 1 \times P(X=1) + 2 \times P(X=2)$
$E(X) = \dfrac{16N + 2N(N-1)}{(N+8)(N+7)} = \dfrac{2N(N+7)}{(N+8)(N+7)} = \dfrac{2N}{N+8}$
Given $E(X) = 1.2$, we have:
$\dfrac{2N}{N+8} = 1.2$
$2N = 1.2N + 9.6$
$0.8N = 9.6$
$N = 12$
Answer: $12$
Hence, the correct answer is option: 'C'.